awulz wrote:For the two textbook problems, we get a root (r) that is repeated twice. (m=2) Following our procedure, we would have matrix A as:

A = [x1 x2; x3 x4] where x1, x2, x3, and x4 are integers.

We now must find A*A. If A*A = [0 0; 0 0], is the generalized eigenvector always [1; 0]? From what I can see, it could be any integer, and therefore, there could be infinite solutions? Or can we not apply our method to the 2x2.

Thanks in advance.

Do you mean (A - lambda I)^2 = 0? If so, then the reduced matrix is just all zeros and the two solutions are [1 0] and [0 1].

There are an infinite number of solutions and there always will be. In the case of just a regular eigenvalue, you can scale it arbitrarily, so there are an infinite number of solutions. In the case were the dimension of the null space of the generalized eigenspace is greater than one, you need to find a basis. For matrices of real numbers, there will always be an infinite number of choices for this. The procedure that I gave in class is just one way to find a set of solutions.